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A two dimensional flow has velocities in x and y directions given by u = 2xyt and v =−y2t , where t denotes time. The equation for streamline passing through x = 1, y = 1 is
- a)x2y = 1
- b)x2y2 = 1
- c)xy2 = 1
- d)x/y2 = 1
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A two dimensional flow has velocities in x and y directions given by u...
Given : u = 2xyt and v = −y2t
On integrating
At x = 1 , y = 1
xy2 = 1
Hence, the correct option is (C).
On integrating
At x = 1 , y = 1
xy2 = 1
Hence, the correct option is (C).
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Community Answer
A two dimensional flow has velocities in x and y directions given by u...
2x^2, where t is time. The flow field represents the motion of fluid particles in a two-dimensional space.
To find the equation for the pathline of a fluid particle, we need to solve the system of differential equations for x and y:
dx/dt = u = 2xyt
dy/dt = v = 2x^2
Let's start by integrating the first equation with respect to t:
∫ dx = ∫ 2xyt dt
Integrating both sides gives:
x = xyt^2 + C1
where C1 is the constant of integration.
Next, let's integrate the second equation with respect to t:
∫ dy = ∫ 2x^2 dt
Integrating both sides gives:
y = (2/3)x^2t + C2
where C2 is the constant of integration.
Now we have the equations for x and y in terms of t. By eliminating t, we can find the equation for the pathline.
Rearranging the first equation, we have:
t^2 = (x - C1) / (xy)
Substituting this into the second equation gives:
y = (2/3)x^2((x - C1) / (xy)) + C2
Simplifying this equation gives:
3y = 2x(x - C1) + 3C2
Expanding and rearranging terms gives:
2x^2 - 2C1x + 3y - 3C2 = 0
This is the equation for the pathline of a fluid particle in the two-dimensional flow.
To find the equation for the pathline of a fluid particle, we need to solve the system of differential equations for x and y:
dx/dt = u = 2xyt
dy/dt = v = 2x^2
Let's start by integrating the first equation with respect to t:
∫ dx = ∫ 2xyt dt
Integrating both sides gives:
x = xyt^2 + C1
where C1 is the constant of integration.
Next, let's integrate the second equation with respect to t:
∫ dy = ∫ 2x^2 dt
Integrating both sides gives:
y = (2/3)x^2t + C2
where C2 is the constant of integration.
Now we have the equations for x and y in terms of t. By eliminating t, we can find the equation for the pathline.
Rearranging the first equation, we have:
t^2 = (x - C1) / (xy)
Substituting this into the second equation gives:
y = (2/3)x^2((x - C1) / (xy)) + C2
Simplifying this equation gives:
3y = 2x(x - C1) + 3C2
Expanding and rearranging terms gives:
2x^2 - 2C1x + 3y - 3C2 = 0
This is the equation for the pathline of a fluid particle in the two-dimensional flow.
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A two dimensional flow has velocities in x and y directions given by u = 2xyt and v =−y2t , where t denotes time. The equation for streamline passing through x = 1, y = 1 isa)x2y = 1b)x2y2 = 1c)xy2 = 1d)x/y2 = 1Correct answer is option 'C'. Can you explain this answer?
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